Write equations for the reaction of propanoic acid with methanol and name any organic products.

This reaction is the formation of an ester because the reactants are a carboxylic acid and an alcohol.
Propanoic acid has 3 carbons and will have the functional group COOH:
CH₃CH₂COOH
Methanol has 1 carbon and an OH functional group:
CH₃OH
So the equation is:
CH₃CH₂COOH + CH₃OH ---> CH₃CH₂COOCH₃ + H₂O
The organic product is methyl propanoate because there are 3 carbons before the ester group and 1 carbon after the ester group.